Chapter 16 Section 3

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# Chapter 16 Section 3 - PowerPoint PPT Presentation

Chapter 16 Section 3. Behavior of Gases. Pressure. Gas particles are constantly moving and colliding, which results in pressure (Force/area) Containers (balloons, tires) remain inflated because particles collide with the walls of the container

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### Chapter 16 Section 3

Behavior of Gases

Pressure
• Gas particles are constantly moving and colliding, which results in pressure (Force/area)
• Containers (balloons, tires) remain inflated because particles collide with the walls of the container
• If more particles of gas are pumped into the container, there will be more collisions and the walls will be pushed further outward
Pascal
• Pressure is measured in pascals (Pa)
• 1 Pa = 1 N/1m2
• 1000 Pa = 1 kPa
• At Earth’s surface, the atmosperic pressure = 101.3 kPa
• 101,300 N per square meter
Boyle’s Law
• What happens to gas pressure if you decrease the size of the container?
• Particles will strike each other and the walls more often, increasing pressure
• If you give the gas particles more space, they will hit the walls less often, pressure will decrease
• Weather balloons
Boyle’s Law in Action (Volume-Pressure Equation)
• A balloon has a volume of 10.0 L at a pressure of 101 kPa. What will be the new volume when the pressure drops to 43.0 kPa?
• P1V1 = P2V2
• 101 x 10.0 = 43.0 x V2
• 1010 = 43.0 x V2

43.0 43.0

• 23.5 L = V2
The Pressure-Temperature Relationship
• Why do you need to keep pressurized spray canisters away from heat?
• Hotter temp. = faster moving particles (more collisions with the walls)
• Volume can’t be increased (rigid canister)
• Pressure increases
• Canister will explode
Charles’s Law
• Gases expand when they are heated (hot air balloons)
• Hot air is less dense than cool air
• The volume of gas increases with increasing temp.

(also, the volume of gas decreases with decreasing temp)

• Gas is heated  particles move faster  particles strike the walls of their container more often and with more force  Larger volume
Using Charles’s Law
• Temperature must be in Kelvin
• V1 = V2

T1 T2

• What would be the resulting volume of a 2.0 L balloon at 25 C that was placed in a container with ice water at 3 C ?
• T1 = 25 C + 273 = 298 K
• T2 = 3 C + 273 = 276 K
• 2 = V2

298 276

• 2 x 276 = 298 x V2
• 552 = 298 x V2
• 1.9 L = V2